Digital Logic & Number representation

Question 1
In the following truth table, V = 1 if and only if the input is valid. gatecs20133 What function does the truth table represent?
Tick
Priority encoder
Cross
Decoder
Cross
Multiplexer
Cross
Demultiplexer


Question 1-Explanation: 
Since there are more than one outputs and number of outputs is less than inputs, it is a Priority encoder V=1 when input is valid and for priority encoder it checks first high bit encountered. Except all are having at least one bit high and ‘x’ represents the “don’t care” as we have found a high bit already. So answer is (A).
Question 2
Which one of the following expressions does NOT represent exclusive NOR of x and y?
Cross
xy+x\'y\'
Cross
x⊕y\'
Cross
x\'⊕y
Tick
x\'⊕y\'


Question 2-Explanation: 
By Definition of XNOR, x\\odot y = x\' y\' + xy So Option-A is correct. Also by Definition of XOR, x\\oplus y = x\' y + xy\' Option-B is x\\oplus y\' = x\' y\' + x(y\')\' = x\' y\' + xy = x\\odot y So Option-B is also correct. Option-C is x\' \\oplus y = (x\')\' y + x\' y\' = x\' y\' + xy = x\\odot y Option-C is also correct. Option-D x\'⊕y\' = x\'\'y\' + x\'y\'\' = xy\' + x\'y = x⊕y ≠ x⊙y Therefore option (D) is false. This explanation is provided by Chirag Manwani.
Question 3
The truth table truthtable represents the Boolean function
Tick
X
Cross
X+Y
Cross
X xor Y
Cross
Y


Question 3-Explanation: 
The value of f(X, Y) is same as X for all input pairs. Also sum of product form of expression we get,
= XY’+XY
= X(Y’+Y)
= X *1
= X 
We see from truth table –
Column x = f(x,y) 
So , 
f(x,y)=x 
Option (A) is correct.
Question 4

What is the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term.

 

Cross

b\'d\'

Tick

b\'d\' + b\'c\'

Cross

b\'d\' + a\'b\'c\'d\'

Cross

b\'d\' + b\'c\' + c\'d\'



Question 4-Explanation: 

There are two prime implicants in the following K-Map- 

 

Prime Implicant highlighted in Green = b\'c\'

Prime Implicant highlighted in Orange = b\'d\'

So the Boolean expression is- b\'c\' + b\'d\'

Therefore option (B) is correct. 

Question 5

What is the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term.

 

Cross

b\'d\'

Tick

b\'d\' + b\'c\'

Cross

b\'d\' + a\'b\'c\'d\'

Cross

b\'d\' + b\'c\' + c\'d\'



Question 5-Explanation: 

There are two prime implicants in the following K-Map- 

 

Prime Implicant highlighted in Green = b\'c\'

Prime Implicant highlighted in Orange = b\'d\'

So the Boolean expression is- b\'c\' + b\'d\'

Therefore option (B) is correct. 

Question 6
Which one of the following circuits is NOT equivalent to a 2-input XNOR (exclusive NOR) gate? GATECS201113
Cross
A
Cross
B
Cross
C
Tick
D


Question 6-Explanation: 
All options except D produce XOR as described below :  NOT({a \\bigoplus b}) = XNOR \\\\ NOT(NOT(a)\\bigoplus NOT(b)) = NOT(aNOT(b) + NOT(a)b) = XNOR \\\\ NOT(a) \\bigoplus b = ab + NOT(ab)= XNOR \\\\ NOT(NOT(aNOT(b))(NOT(b)+a)) = NOT(NOT(ab)+ba) = NOT(XNOR)
Question 7
The simplified SOP (Sum Of Product) form of the boolean expression (P + Q' + R') . (P + Q' + R) . (P + Q + R') is
Cross
(P\'.Q + R\')
Tick
(P + Q\'.R\')
Cross
(P\'.Q + R)
Cross
(P.Q + R)


Question 7-Explanation: 
See following : (P+Q\'+R\').(P+Q\'+R).(P+Q+R\') = \\prod(3, 2, 1) = \\sum(0, 4, 5, 6, 7) \"gate2011A24\" From the K-map, POS form is : P + Q\'.R\'
Question 8

Consider the following circuit involving three D-type flip-flops used in a certain type of counter configuration. 

 

If at some instance prior to the occurrence of the clock edge, P, Q and R have a value 0, 1 and 0 respectively, what shall be the value of PQR after the clock edge?

Cross

000

Cross

001

Cross

010

Tick

011



Question 8-Explanation: 

P\' = R Q\' = (P + R)\' R\' = QR\' Given that (P, Q, R) = (0, 1, 0), next state P\', Q\', R\' = 0, 1, 1 ----------------------------------------------------------------------------------------------- D flip flop truth table

DQ(t+1)
00
11

Initially (p,q,r) =(0,1,0) D for p=R D for q=NOT(p xor r) D for r= (not)r.q So Q(t+1) for(p,q,r) p=>r=0 so p=0 q=> NOT(p xor r) => 1      so q=1 r=>(not)r.q => 1         so r=1 (p, q, r) = (0, 1, 1) Alternative approach - Truth table of a D Flip-Flop- \\begin{tabular}{|c|c|} \\hline D_{t} & Q_{t+1}\\\\ \\hline 0 & 0\\\\ \\hline 1 & 1\\\\ \\hline \\end{tabular}  By looking at the circuit diagram, it is clear that the boolean expressions of P, Q, and R are- Here the subscript t refers to the current clock cycle, and the subscript (t+1) refers to the next clock cycle. Q_{P(t+1)} = P_{t+1} = R_t  [Tex]Q_{Q(t+1)} = Q_{t+1} = R_{t}\' P_{t}\'  [/Tex]Q_{R(t+1)} = R_{t+1} = Q_{t} R_{t}\'  [Tex]\\begin{tabular}{|c|c|c|} \\hline D_{t} & Q_{t+1}\\\\ \\hline 0 & 0\\\\ \\hline 1 & 1\\\\ \\hline \\end{tabular}  [/Tex]This explanation is provided by Chirag Manwani

Question 9

Consider the data given in previous question. If all the flip-flops were reset to 0 at power on, what is the total number of distinct outputs (states) represented by PQR generated by the counter?

Cross

3

Tick

4

Cross

5

Cross

6



Question 9-Explanation: 

There are four distinct states, 000 → 010 → 011 → 100 (→ 000) so the answer is B

Question 10
The minterm expansion of f(P, Q, R) = PQ + QR' + PR' is
Tick
m2 + m4 + m6 + m7
Cross
m0 + m1 + m3 + m5
Cross
m0 + m1 + m6 + m7
Cross
m2 + m3 + m4 + m5


Question 10-Explanation: 
K-map,
= PQ + QR’ + PR’ 
= PQ(R+R’) + (P+P’)QR’ + P(Q+Q’)R’
= PQR + PQR’ +PQR’ +P’QR’ + PQR’ + PQ’R’ 
= PQR(m7) + PQR\'(m6)+P’QR\'(m2) +PQ’R\'(m4) 
= m2 + m4 + m6 + m7 
Option (A) is correct.
There are 267 questions to complete.

  • Last Updated : 27 Sep, 2021

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